A081720 -id:A081720 - cp's OEIS frontend

A208544 T(n,k) = Number of n-bead necklaces of k colors allowing reversal, with no adjacent beads having the same color.

1, 2, 0, 3, 1, 0, 4, 3, 0, 0, 5, 6, 1, 1, 0, 6, 10, 4, 6, 0, 0, 7, 15, 10, 21, 3, 1, 0, 8, 21, 20, 55, 24, 13, 0, 0, 9, 28, 35, 120, 102, 92, 9, 1, 0, 10, 36, 56, 231, 312, 430, 156, 30, 0, 0, 11, 45, 84, 406, 777, 1505, 1170, 498, 29, 1, 0, 12, 55, 120, 666, 1680, 4291, 5580, 4435

Offset: 1

R. H. Hardin, Feb 27 2012

Table starts
.1.2..3...4....5.....6......7......8.......9......10......11.......12.......13
.0.1..3...6...10....15.....21.....28......36......45......55.......66.......78
.0.0..1...4...10....20.....35.....56......84.....120.....165......220......286
.0.1..6..21...55...120....231....406.....666....1035....1540.....2211.....3081
.0.0..3..24..102...312....777...1680....3276....5904....9999....16104....24882
.0.1.13..92..430..1505...4291..10528...23052...46185...86185...151756...254618
.0.0..9.156.1170..5580..19995..58824..149796..341640..714285..1391940..2559414
.0.1.30.498.4435.25395.107331.365260.1058058.2707245.6278140.13442286.26942565

			All solutions for n=7, k=3:
..1....1....1....1....1....1....1....1....1
..2....2....2....2....2....2....2....2....2
..3....3....1....1....3....1....3....1....3
..1....1....2....2....1....2....2....3....2
..2....3....3....3....3....1....3....1....3
..3....1....1....2....2....2....2....2....1
..2....3....3....3....3....3....3....3....3

Andrew Howroyd, Table of n, a(n) for n = 1..1275 (first 264 terms from R. H. Hardin)

Cf. A081720, A208535, A106512.
Main diagonal is A208538.
Columns 3..7 are A208539, A208540, A208541, A208542, A208543.
Row 2 is A000217(n-1).
Row 3 is A000292(n-2).
Row 4 is A002817(n-1).
Row 5 is A164938(n-1).
Row 6 is A027670(n-1).

T[n_, k_] := If[n == 1, k, (DivisorSum[n, EulerPhi[n/#]*(k-1)^#&]/n + If[ OddQ[n], 1-k, k*(k-1)^(n/2)/2])/2]; Table[T[n-k+1, k], {n, 1, 12}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Oct 30 2017, after Andrew Howroyd *)

T(n, k) = if(n==1, k, (sumdiv(n, d, eulerphi(n/d)*(k-1)^d)/n + if(n%2, 1-k, k*(k-1)^(n/2)/2))/2);
for(n=1, 10, for(k=1, 10, print1(T(n, k), ", ")); print) \\ Andrew Howroyd, Oct 14 2017

T(2n+1,k) = A208535(2n+1,k)/2 for n > 0, T(2n,k) = (A208535(2n,k) + (k*(k-1)^n)/2)/2. - Andrew Howroyd, Mar 12 2017
Empirical for row n:
n=1: a(k) = k
n=2: a(k) = (1/2)*k^2 - (1/2)*k
n=3: a(k) = (1/6)*k^3 - (1/2)*k^2 + (1/3)*k
n=4: a(k) = (1/8)*k^4 - (1/4)*k^3 + (3/8)*k^2 - (1/4)*k
n=5: a(k) = (1/10)*k^5 - (1/2)*k^4 + k^3 - k^2 + (2/5)*k
n=6: a(k) = (1/12)*k^6 - (1/2)*k^5 + (3/2)*k^4 - (7/3)*k^3 + (23/12)*k^2 - (2/3)*k
n=7: a(k) = (1/14)*k^7 - (1/2)*k^6 + (3/2)*k^5 - (5/2)*k^4 + (5/2)*k^3 - (3/2)*k^2 + (3/7)*k

A032276 Number of bracelets (turnover necklaces) with n beads of 5 colors.

Original entry on oeis.org

5, 15, 35, 120, 377, 1505, 5895, 25395, 110085, 493131, 2227275, 10196680, 46989185, 218102685, 1017448143, 4768969770, 22440372245, 105966797755, 501938733555, 2384200683816, 11353290089305

Offset: 1

Christian G. Bower

nonn

From Petros Hadjicostas, Sep 01 2018: (Start)
The DIK transform of the sequence (c(n): n >= 1), with g.f. C(x) = Sum_{n >= 1} c(n)*x^n, has g.f. -(1/2)*Sum_{m >= 1} (phi(m)/m))*log(1-C(x^m)) + (1 + C(x))^2/(4*(1-C(x^2))) - 1/4.
Here, c(1) = 5 and c(n) = 0 for n >= 2, and thus, C(x) = 5*x. Substituting this to the above g.f., we get that the g.f. of the current sequence is A(x) = Sum_{n >= 1} a(n)*x^n = -(1/2)*Sum_{m >= 1} (phi(m)/m))*log(1-5*x^m) + (1 + 5*x)^2/(4*(1-5*x^2)) - 1/4. This agrees with Herbert Kociemba's g.f. below except for an extra 1 because (1 + (1+5*x+10*x^2)/(1-5*x^2))/2 = 1 + (1 + 5*x)^2/(4*(1-5*x^2)) - 1/4.
(End)

			For n=2, the 15 bracelets are AA, AB, AC, AD, AE, BB, BC, BD, BE, CC, CD, CE, DD, DE, and EE. - _Robert A. Russell_, Sep 24 2018

C. G. Bower, Transforms (2)
F. Ruskey, Necklaces, Lyndon words, De Bruijn sequences, etc.
F. Ruskey, Necklaces, Lyndon words, De Bruijn sequences, etc. [Cached copy, with permission, pdf format only]
M. Taniguchi, H. Du, and J. S. Lindsey, Enumeration of virtual libraries of combinatorial modular macrocyclic (bracelet, necklace) architectures and their linear counterparts, Journal of Chemical Information and Modeling, 53 (2013), 2203-2216.
Index entries for sequences related to bracelets

Cf. A081720.
Column 5 of A051137.
Cf. A001869 (oriented), A056487 (achiral), A278641 (chiral).

mx=40;CoefficientList[Series[(1-Sum[ EulerPhi[n]*Log[1-5*x^n]/n,{n,mx}]+(1+5 x+10 x^2)/(1-5 x^2))/2,{x,0,mx}],x] (* Herbert Kociemba, Nov 02 2016 *)
k=5; Table[DivisorSum[n, EulerPhi[#] k^(n/#) &]/(2n) + (k^Floor[(n+1)/2] + k^Ceiling[(n+1)/2])/4, {n, 1, 30}] (* Robert A. Russell, Sep 24 2018 *)

"DIK" (bracelet, indistinct, unlabeled) transform of 5, 0, 0, 0, ...
a(n) = A081720(n,5), n >= 1. - Wolfdieter Lang, Jun 03 2012
G.f.: (1 - Sum_{n>=1} phi(n)*log(1 - 5*x^n)/n + (1+5*x+10*x^2)/(1-5*x^2))/2. - Herbert Kociemba, Nov 02 2016
a(n) = (3/2)*5^(n/2) + (1/(2*n))*Sum_{d|n} phi(n/d)*5^d, if n is even, and = (1/2)*5^((n+1)/2) + (1/(2*n))*Sum_{d|n} phi(n/d)*5^d, if n is odd. - Petros Hadjicostas, Sep 01 2018
a(n) = (A001869(n) + A056487(n+1)) / 2 = A278641(n) + A056487(n+1) = A001869(n) - A278641(n). - Robert A. Russell, Oct 13 2018
a(n) = (k^floor((n+1)/2) + k^ceiling((n+1)/2))/4 + (1/(2*n))*Sum_{d divides n} phi(d)*k^(n/d), where k=5 is the maximum number of colors. - Richard L. Ollerton, May 04 2021
a(n) = (k^floor((n+1)/2) + k^ceiling((n+1)/2))/4 + (1/(2*n))*Sum_{i=1..n} k^gcd(n,i), where k=5 is the maximum number of colors. (See A051137.) - Richard L. Ollerton, May 04 2021

A276550 Array read by antidiagonals: T(n,k) = number of primitive (period n) bracelets using a maximum of k different colored beads.

Original entry on oeis.org

1, 2, 0, 3, 1, 0, 4, 3, 2, 0, 5, 6, 7, 3, 0, 6, 10, 16, 15, 6, 0, 7, 15, 30, 45, 36, 8, 0, 8, 21, 50, 105, 132, 79, 16, 0, 9, 28, 77, 210, 372, 404, 195, 24, 0, 10, 36, 112, 378, 882, 1460, 1296, 477, 42, 0, 11, 45, 156, 630, 1848, 4220, 5890, 4380, 1209, 69, 0

Offset: 1

Andrew Howroyd, Apr 09 2017

Turning over will not create a new bracelet.

			Table starts:
  1  2   3    4     5      6      7       8 ...
  0  1   3    6    10     15     21      28 ...
  0  2   7   16    30     50     77     112 ...
  0  3  15   45   105    210    378     630 ...
  0  6  36  132   372    882   1848    3528 ...
  0  8  79  404  1460   4220  10423   22904 ...
  0 16 195 1296  5890  20640  60021  151840 ...
  0 24 477 4380 25275 107100 364854 1057392 ...
  ...

Andrew Howroyd, Table of n, a(n) for n = 1..1275
G. Melançon, C. Reutenauer, On a Class of Lyndon Words Extending Christoffel Words and Related to a Multidimensional Continued Fraction Algorithm, J. Int. Seq. 16 (2013) #13.9.7, Corollary 6.
F. Ruskey, Necklaces, Lyndon words, De Bruijn sequences, etc.
F. Ruskey, Necklaces, Lyndon words, De Bruijn sequences, etc. [Cached copy, with permission, pdf format only]

Columns 2-6 are A001371, A032294, A032295, A032296, A056347.

Cf. A081720, A273891, A276543, A152176.

A276550 := proc(n,k)
    local d ;
    add( numtheory[mobius](n/d)*A081720(d,k),d=numtheory[divisors](n)) ;
end proc:
seq(seq(A276550(n,d-n),n=1..d-1),d=2..10) ; # R. J. Mathar, Jan 22 2022

t[n_, k_] := Sum[EulerPhi[d] k^(n/d), {d, Divisors[n]}]/(2n) + (k^Floor[(n+1)/2] + k^Ceiling[(n+1)/2])/4;
T[n_, k_] := Sum[MoebiusMu[d] t[n/d, k], {d, Divisors[n]}];
Table[T[n-k+1, k], {n, 1, 11}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Mar 26 2020 *)

T(n, k) = Sum_{d|n} mu(n/d) * A081720(d,k) for k<=n. Corrected Jan 22 2022

A051137 Table T(n,k) read by antidiagonals: number of necklaces allowing turnovers (bracelets) with n beads of k colors.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 6, 10, 10, 5, 1, 1, 8, 21, 20, 15, 6, 1, 1, 13, 39, 55, 35, 21, 7, 1, 1, 18, 92, 136, 120, 56, 28, 8, 1, 1, 30, 198, 430, 377, 231, 84, 36, 9, 1, 1, 46, 498, 1300, 1505, 888, 406, 120, 45, 10, 1

Offset: 0

Alford Arnold

Unlike A075195 and A284855, antidiagonals go from bottom-left to top-right.

			Table begins with T[0,1]:
1  1    1     1      1       1        1        1         1         1
1  2    3     4      5       6        7        8         9        10
1  3    6    10     15      21       28       36        45        55
1  4   10    20     35      56       84      120       165       220
1  6   21    55    120     231      406      666      1035      1540
1  8   39   136    377     888     1855     3536      6273     10504
1 13   92   430   1505    4291    10528    23052     46185     86185
1 18  198  1300   5895   20646    60028   151848    344925    719290
1 30  498  4435  25395  107331   365260  1058058   2707245   6278140
1 46 1219 15084 110085  563786  2250311  7472984  21552969  55605670
1 78 3210 53764 493131 3037314 14158228 53762472 174489813 500280022

C. G. Bower, Transforms (2)

Columns 2-6 are A000029, A027671, A032275, A032276, and A056341.
Rows 2-7 are A000217, A000292, A002817, A060446, A027670, and A060532.
Cf. A000031.
Cf. A081720, A081721.
T(n,k) = (A075195(n,k) + A284855(n,k)) / 2.

b[n_, k_] := DivisorSum[n, EulerPhi[#]*k^(n/#) &] / n;
c[n_, k_] := If[EvenQ[n], (k^(n/2) + k^(n/2+1))/2, k^((n+1)/2)];
T[0, ] = 1; T[n, k_] := (b[n, k] + c[n, k])/2;
Table[T[n, k-n], {k, 1, 11}, {n, k-1, 0, -1}] // Flatten
(* Robert A. Russell, Sep 21 2018 after Jean-François Alcover *)

T(n, k) = (k^floor((n+1)/2) + k^ceiling((n+1)/2)) / 4 + (1/(2*n)) * Sum_{d divides n} phi(d) * k^(n/d). - Robert A. Russell, Sep 21 2018
G.f. for column k: (kx/4)*(kx+x+2)/(1-kx^2) - Sum_{d>0} phi(d)*log(1-kx^d)/2d. - Robert A. Russell, Sep 28 2018
T(n, k) = (k^floor((n+1)/2) + k^ceiling((n+1)/2))/4 + (1/(2*n))*Sum_{i=1..n} k^gcd(n,i). (See A075195 formulas.) - Richard L. Ollerton, May 04 2021

A056341 Number of bracelets of length n using a maximum of six different colored beads.

Original entry on oeis.org

6, 21, 56, 231, 888, 4291, 20646, 107331, 563786, 3037314, 16514106, 90782986, 502474356, 2799220041, 15673673176, 88162676511, 497847963696, 2821127825971, 16035812864946, 91404068329560

Offset: 1

Marks R. Nester

nonn

Turning over will not create a new bracelet.

			For n=2, the 21 bracelets are AA, AB, AC, AD, AE, AF, BB, BC, BD, BE, BF, CC, CD, CE, CF, DD, DE, DF, EE, EF, and FF. - _Robert A. Russell_, Sep 24 2018

M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for pdf file of Chap. 2.]

Cf. A000029, A054625.
Cf. a(n) = A081720(n,6), n >= 6. - Wolfdieter Lang, Jun 03 2012
Column 6 of A051137.
Equals (A054625 + A056488) / 2 = A054625 - A278642 = A278642 + A056488.

mx=40;CoefficientList[Series[(1-Sum[ EulerPhi[n]*Log[1-6*x^n]/n,{n,mx}]+(1+6 x+15 x^2)/(1-6 x^2))/2,{x,0,mx}],x] (* Herbert Kociemba, Nov 02 2016 *)
k=6; Table[DivisorSum[n, EulerPhi[#] k^(n/#) &]/(2n) + (k^Floor[(n+1)/2] + k^Ceiling[(n+1)/2])/4, {n, 1, 30}] (* Robert A. Russell, Sep 24 2018 *)

a(n) = Sum_{d|n} phi(d)*6^(n/d)/(2*n);
a(n) = 6^((n+1)/2)/2 for n odd,
       (7/4)*6^(n/2) for n even.
G.f.: (1 - Sum_{n>=1} phi(n)*log(1 - 6*x^n)/n + (1+6*x+15*x^2)/(1-6*x^2))/2. - Herbert Kociemba, Nov 02 2016

A214306 Triangle with entry a(n,m) giving the total number of bracelets of n beads (D_n symmetry) with n colors available for each bead, but only m distinct colors present, with m from {1, 2, ..., n} and n >= 1.

Original entry on oeis.org

1, 2, 1, 3, 6, 1, 4, 24, 24, 3, 5, 60, 180, 120, 12, 6, 165, 1120, 2040, 900, 60, 7, 336, 5145, 21420, 25200, 7560, 360, 8, 784, 23016, 183330, 442680, 335160, 70560, 2520, 9, 1584, 91056, 1320480, 5846400, 8890560, 4656960, 725760, 20160, 10

Offset: 1

Wolfdieter Lang, Jul 20 2012

This triangle is obtained from the array A213941 by summing in row n, for n >= 1, all entries related to partitions of n with the same number of parts m.
a(n,m) is the total number of necklaces of n beads (dihedral D_n symmetry) corresponding to all the color multinomials obtained from all p(n,m) = A008284(n,m) partitions of n with m parts, written in nonincreasing form, by 'exponentiation'. Therefore only m from the available n colors are present, and a(n,m) gives the number of bracelets with n beads with only m of the n available colors present, for m from 1,2,...,n, and n >= 1. All of the possible color assignments are counted.
See the comments on A212359 for the Abramowitz-Stegun (A-St) order of partitions, and the 'exponentiation' to obtain multisets, used to encode color multinomials, from partitions. See a link in A213938 for representative multisets for given signature used to define color multinomials.
The row sums of this triangle coincide with the ones of array A213941, and they are given by A081721.

			n\m 1    2     3       4       5       8       7      8     9
1   1
2   2    1
3   3    6     1
4   4   24    24       3
5   5   60   180     120      12
6   6  165  1120    2040     900     60
7   7  336  5145   21420   25200   7560      360
8   8  784 23016  183330  442680  33516    70560   2520
9   9 1584 91056 1320480 5846400 8890560 4656960 725760 20160
...
Row n=10:  10, 3420, 357480, 8691480, 64420272, 172609920, 177811200, 68040000, 8164800, 181440;
Row n=11:  11, 6820, 1327095, 52727400, 622175400, 2714009760, 4837417200, 3592512000, 1047816000, 99792000, 1814400.
a(2,2) = 1 from the color monomial c[1]^1*c[2]^1 = c[1]*c[2] (from the m=2 partition [1,1] of n=2). The bracelet in question is cyclic(12) (we use j for color c[j] in these examples). The same holds for the necklace case.
a(5,3) = 60 + 120 = 180, from A213941(5,4) + A213941(5,5), because k(5,3,1) = A214314(5,3) = 4 and p(5,3)=2.
a(3,1) = 3 from the color monomials c[1]^3, c[2]^3 and c[3]^3. The three bracelets are cyclic(111), cyclic(222) and cyclic(333). The same holds for the necklace case.
In general a(n,1)=n from the partition [n] providing the color signature (exponent), and the n color choices.
a(3,2) = 6 from the color signature c[.]^2 c[.]^1, (from the m=2 partition [2,1] of n=3), and there are 6 choices for the color indices. The 6 bracelets are cyclic(112), cyclic(113), cyclic(221), cyclic(223), cyclic(331) and cyclic(332). The same holds for the necklace case.
a(3,3) = 1. The color multinomial is c[1]*c[2]*c[3] (from the m=3 partition [1,1,1]). All three available colors are used. There is only one bracelet: cyclic(1,2,3). The necklace cyclic(1,3,2) becomes equivalent under D_3 operation.
a(4,2) = 24 from two color signatures c[.]^3 c[.] and c[.]^2 c[.]^2 (from the two m=2 partitions of n=4: [3,1] and [2,2]). The first one produces 4*3=12 bracelets, namely 1112, 1113, 1114, 2221, 2223, 2224, 3331, 3332, 3334, 4441, 4442 and 4443, all taken cyclically. The second color signature leads to another 6*2=12 bracelets: 1122, 1133, 1144, 2233, 2244, 3344, 1212, 1313, 1414, 2323, 2424 and 3434, all taken cyclically. Together they provide the 24 bracelets counted by a(4,2). The same holds for the necklace case.
a(4,3) = 24 from the color signature c[.]^2 c[.]c[.]. There are 4*3 =12 color choices each with two bracelets: 1123, 1213, 1124, 1214, 1134, 1314, 2213, 2123, 2214, 2124, 2234, 2324, 3312, 3132, 3314, 3134, 3324, 3234, 4412, 4142, 4413, 4143, 4423 and 4243, each taken cyclically.

Andrew Howroyd, Table of n, a(n) for n = 1..1275

Cf. A081721, A212360 (necklaces), A213941, A273891.

(* t = A081720 *) t[n_, k_] := (For[t1 = 0; d = 1, d <= n, d++, If[Mod[n, d] == 0, t1 = t1 + EulerPhi[d]*k^(n/d)]]; If[EvenQ[n], (t1 + (n/2)*(1 + k)*k^(n/2))/(2*n), (t1 + n*k^((n + 1)/2))/(2*n)]);
T[n_, k_] := Binomial[n, k]*Sum[(-1)^i * Binomial[k, i]*t[n, k - i], {i, 0, k - 1}];
Table[T[n, k], {n, 1, 10}, {k, 1, n}] // Flatten (* Jean-François Alcover, Oct 08 2017, after Andrew Howroyd *)

a(n,m) = Sum_{j=1..p(n,m)} A213941(n, k(n,m,1)+j-1), with k(n,m,1) = A214314(n,m) the position where in the list of partitions of n in A-St order the first with m parts appears, and p(n,m) is the number of partitions of n with m parts shown in the array A008284. E.g., n=5, m=3: k(5,3,1) = A214314(5,3) = 4, p(5,3) = 2.

a(n,m) = binomial(n,m) * A273891(n,m). - Andrew Howroyd, Mar 25 2017

A295925 Number of bilaterally asymmetric 8-hoops with n symbols.

Original entry on oeis.org

6, 336, 3795, 23520, 102795, 355656, 1039626, 2674440, 6223140, 13354440, 26807781, 50885016, 92095185, 159981360, 268161060, 435614256, 688255506, 1060829280, 1599170055, 2362871280, 3428409831, 4892775096, 6877654350

Offset: 2

Petros Hadjicostas, Nov 30 2017

This is a corrected version of the entries in sequence A210769, which is copied from the third row of Table 2 (p. 381) in Williamson (1972). Apparently, in that row, there are probable typographical errors for the values of a(4) and a(7). The formula for a(n) can be obtained by letting z_1=z_2=...=z_8=n in equation (24) on p. 377 in Williamson (1972). In any case, to be certain, we provide a sketch of an independent derivation of the formula.
Bilaterally symmetric bracelets are also known as circular palindromes. This kind of necklaces was first studied by Sommerville (1909) in the context of circular compositions.
Consider sequence A081720, which contains the numbers T(n,k) that are the number of bracelets (turn over necklaces) with n beads each of which is colored with one of k colors. The g.f. for column k of that triangle is (1/2)*((k*x+k*(k+1)*x^2/2)/(1-k*x^2) - Sum_{n>=1} (phi(n)/n)*log(1-k*x^n)). The part of the g.f. that is the number of bilaterally symmetric bracelets with n beads of k colors is (k*x+k*(k+1)*x^2/2)/(1-k*x^2). Thus, for fixed k (= number of colors for sequence A081720), the g.f. of the number of bilaterally asymmetric bracelets with n beads of k colors is the difference between the two g.f.'s, that is, (1/2)*(-(k*x+k*(k+1)*x^2/2)/(1-k*x^2) - Sum_{n>=1} (phi(n)/n)*log(1-k*x^n)). The coefficient of x^8 in the Taylor expansion w.r.t. x (around x=0) for the latter g.f. gives the number of bilaterally asymmetric 8-hoops obtained using (up to) k symbols. Taking the 8th derivative w.r.t. x of the last g.f., evaluating at x=0, dividing by 8!, and replacing k with n, we get the formulae given below.

			From the A060560(2) = 30 8-hoops (i.e., from the total number of ways of coloring the vertices of an octagon using up to n=2 colors, allowing for rotations and reflections), there are A019583(2+1) = 24 that are circular palindromes (i.e., bilaterally symmetric bracelets). Hence, there are 30-24=6 bilaterally asymmetric 8-hoops using up to 2 colors. They are the following: 01001111, 01000111, 01000011, 00101011, 00110111 and 11001000. (To view these 6 asymmetric bracelets, the 0's and 1's must be placed on the vertices of a regular octagon inscribed in a circle as it is done in Fig. 4 on p. 379 in Williamson (1972), where 0 is replaced by a and 1 by b.)

D. M. Y. Sommerville, On certain periodic properties of cyclic compositions of numbers, Proc. London Math. Soc. S2-7(1) (1909), 263-313.
S. G. Williamson, The combinatorial analysis of patterns and the principle of inclusion-exclusion, Discrete Math. 1 (1972), 357-388.
Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

Cf. A019583, A054622, A060560, A081720, A119963, A210699, A210766, A210768, A295925, A292200.

Drop[#, 2] &@ CoefficientList[Series[3 (7 x^4 + 82 x^3 + 237 x^2 + 92 x + 2) (x + 1) x^2/(1 - x)^9, {x, 0, 24}], x] (* Michael De Vlieger, Dec 02 2017 *)

a(n) = (1/16)*(n^3-n^2-2)*(n^2+n+2)*(n+1)*(n-1)*n = (n^8-4*n^5-3*n^4+2*n^2+4*n)/16.
a(n) = A060560(n) - A019583(n+1) = (A054622(n) - A019583(n+1))/2. (Notice that the offsets of the sequences in these formulae are not necessarily the same as the offset of the current sequence.)
G.f.: 3*(7*x^4 + 82*x^3 + 237*x^2 + 92*x + 2)*(x + 1)*x^2/(1-x)^9.
Recurrence: (1-Delta)^9 a(n) = 0, where Delta^m a(n) = a(n-m). Hence, a(n) = 9*a(n-1)-36*a(n-2)+84*a(n-3)-126*a(n-4)+126*a(n-5)-84*a(n-6)+36*a(n-7)-9*a(n-8)+a(n-9).
E.g.f.: exp(x)*x^2*(48 + 848*x + 1658*x^2 + 1046*x^3 + 266*x^4 + 28*x^5 + x^6)/16. - Stefano Spezia, Feb 18 2024

A060560 Number of ways to color the vertices of an octagon using <= n colors, allowing rotations and reflections.

Original entry on oeis.org

0, 1, 30, 498, 4435, 25395, 107331, 365260, 1058058, 2707245, 6278140, 13442286, 26942565, 51084943, 92383305, 160386360, 268718116, 436365945, 689252778, 1062132490, 1600850055, 2365010571, 3431103775, 4896133188, 6881801550, 9539306725, 13054804776, 17655599430

Offset: 0

N. J. A. Sloane, Apr 12 2001

nonn

In Williamson's terminology, this is "Number of 8-hoops with n symbols."

Harry J. Smith, Table of n, a(n) for n=0..500
S. G. Williamson, The combinatorial analysis of patterns and the principle of inclusion-exclusion, Discrete Math. 1 (1972), no. 4, 357--388. MR0299493 (45 #8541). - _N. J. A. Sloane_, Mar 27 2012
Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

Cf. A081720, A210766, A210768, A295925.

a(n) = { (n^8 + 4*n^5 + 5*n^4 + 2*n^2 + 4*n)/16 } \\ Harry J. Smith, Jul 07 2009

a(n) = n*(n + 1)*(n^6 - n^5 + n^4 + 3*n^3 + 2*n^2 - 2*n + 4)/16.

G.f.: x*(1 + 21*x + 264*x^2 + 949*x^3 + 1014*x^4 + 258*x^5 + 13*x^6)/(1 - x)^9. - Colin Barker, Jan 29 2012

cp's OEIS Frontend

A208544 T(n,k) = Number of n-bead necklaces of k colors allowing reversal, with no adjacent beads having the same color.

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A032276 Number of bracelets (turnover necklaces) with n beads of 5 colors.

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A276550 Array read by antidiagonals: T(n,k) = number of primitive (period n) bracelets using a maximum of k different colored beads.

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A051137 Table T(n,k) read by antidiagonals: number of necklaces allowing turnovers (bracelets) with n beads of k colors.

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A056341 Number of bracelets of length n using a maximum of six different colored beads.

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A214306 Triangle with entry a(n,m) giving the total number of bracelets of n beads (D_n symmetry) with n colors available for each bead, but only m distinct colors present, with m from {1, 2, ..., n} and n >= 1.

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A295925 Number of bilaterally asymmetric 8-hoops with n symbols.

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